Delete smaller nodes in linked list Python program

I tried to write following function to delete the smallest node in a linked list. But, i am getting error and am not able to rectify it. please help.

error is -> Error 1 error C4703: potentially uninitialized local pointer variable 'trailSmall' used

here is my code:

void linkedListType::deleteSmallest[] { nodeType *current; nodeType *trailCurrent; // used for pointing to node just before current nodeType *small; nodeType *trailSmall; if [first == NULL] cout link; while [current != NULL] { if [small->info > current->info] { trailSmall = trailCurrent; small = current; } trailCurrent = current; current = current->link; } if [small == first] first = first->link; else if [small != first] { trailSmall->link = small->link; if [small == last] last = trailSmall; } delete small; } }

Last update on January 04 2021 14:02:46 [UTC/GMT +8 hours]

Write a Python program to delete the last item from a singly linked list.

Sample Solution:-

Python Code:

class Node: # Singly linked node def __init__[self, data=None]: self.data = data self.next = None class singly_linked_list: def __init__[self]: # Createe an empty list self.tail = None self.head = None self.count = 0 def append_item[self, data]: #Append items on the list node = Node[data] if self.head: self.head.next = node self.head = node else: self.tail = node self.head = node self.count += 1 def delete_item[self, data]: # Delete an item from the list current = self.tail prev = self.tail while current: if current.data == data: if current == self.tail: self.tail = current.next else: prev.next = current.next self.count -= 1 return prev = current current = current.next def iterate_item[self]: # Iterate the list. current_item = self.tail while current_item: val = current_item.data current_item = current_item.next yield val items = singly_linked_list[] items.append_item['PHP'] items.append_item['Python'] items.append_item['C#'] items.append_item['C++'] items.append_item['Java'] print["Original list:"] for val in items.iterate_item[]: print[val] print["\nAfter removing the last item from the list:"] items.delete_item['Java'] for val in items.iterate_item[]: print[val]

Sample Output:

Original list: PHP Python C# C++ Java After removing the last item from the list: PHP Python C# C++

Flowchart:

Python Code Editor:

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Returns the items that are parity outliers:

Example:

from collections import Counter def find_parity_outliers[nums]: return [ x for x in nums if x % 2 != Counter[[n % 2 for n in nums]].most_common[][0][0] ] print[find_parity_outliers[[1, 2, 3, 4, 5, 6]]]

Output:

[2, 4, 6]

Delete nodes from the linked list which have a greater value on the right side

In this problem, we have given a singly linked list, and we need to remove all the nodes from it which have a greater value on the right side.

Examples:

• Suppose the list 13 -> 16 -> 9 -> 10 -> 4 -> 6 -> 1 -> 3 -> NULL. This list will be changed to 16 -> 10 -> 6 -> 3 -> NULL. In this, the nodes 13, 9, 4, 1 have been deleted because in linked list there is a greater value on the right side.
• The list 90 -> 80 -> 70 -> 60 -> 50 -> NULL will not be changed because there is no node which has the greater value on the right side.

Method:

We can do this by using the following steps:

• Firstly, we will reverse the given linked list.
• Then, we will traverse the reversed linked list and keep the max node. If the next node is less than the max node, we will delete the next node, otherwise max = next node.
• Finally, we will reverse the linked list again to maintain the order of the linked list given earlier.

C program to delete the nodes which have a greater value on the right side

#include #include struct node { int info; struct node *next; }; struct node *start = NULL; // For inserting the elements in the linked list void add[int item] { struct node *t, *p; t = [struct node *]malloc[ sizeof[ struct node ]]; if[start == NULL] { start = t; start -> info = item; start -> next = NULL; return; } else { struct node *p = start; while[p -> next != NULL] { p = p -> next; } p -> next = t; p = p -> next; p -> info = item; p -> next = NULL; } } // For reversing the nodes of the linked list void reverse [struct node * t] {     struct node * curr = t;     struct node * next = NULL;     struct node * pre = NULL;     while[curr != NULL]     {         next=curr -> next;         curr -> next = pre;         pre = curr;         curr = next;    }    start = pre; } void deleteLesser[struct node * temp] {             reverse[temp]; // For reversing the linked list             delete[start]; // For delete the nodes which have a node with greater value on left side.             reverse[start]; // For reversing the linked list again to maintain the original order of the linked list } // Function for delete the nodes which have a node with greater value on the left side. void delete[struct node * start] {             struct node * curr = start;             struct node * max = start;             struct node * temp;             while[ curr != NULL && curr -> next != NULL]             {                         if[curr -> next -> info < max -> info] // If the curr is smaller than the max, then delete the curr                         {                                     temp = curr -> next;                                     curr -> next = temp -> next;                         }                         else                         {                                     curr = curr -> next;                                     max = curr;                         }             } } // To display the elements of the linked list void traverse[struct node * t] { if[t == NULL] {             printf[" Linked list is empty\n"];                                     }                                     while[t -> next != NULL]                                     {                         printf["%d -> ",t -> info];                         t = t -> next;                         }                         printf["%d\n",t -> info]; } // Driver Function int main[] {     add[13];     add[16];     add[9];     add[10];     add[4];     add[6];     add[1];     add[3];     printf["Given Linked List: \n"];     traverse[start];     deleteLesser[start];     printf["Linked List After deletion: \n"];     traverse[start];     return 0; }

Output:

Time Complexity: The time complexity of the above method is O[n].